Rearrange array such that arr[i] >= arr[j] if i is even and arr[i]<=arr[j] if i is odd and j < i

Difficulty Level : Medium
Last Updated : 14 Nov, 2022

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Given an array of n elements. Our task is to write a program to rearrange the array such that elements at even positions are greater than all elements before it and elements at odd positions are less than all elements before it.
Examples:

Input : arr[] = {1, 2, 3, 4, 5, 6, 7}
Output : 4 5 3 6 2 7 1

Input : arr[] = {1, 2, 1, 4, 5, 6, 8, 8} 
Output : 4 5 2 6 1 8 1 8

The idea to solve this problem is to first create an additional copy of the original array and sort the copied array. Now the total number of even positions in an array with n elements will be floor(n/2) and the remaining is the number of odd positions. Now fill the odd and even positions in the original array using the sorted array in a below manner:

Total odd positions will be n – floor(n/2). Start from (n-floor(n/2))th position in the sorted array and copy the element to the 1st position of the sorted array. Start traversing the sorted array from this position towards the left and keep filling the odd positions in the original array towards the right.
Start traversing the sorted array starting from (n-floor(n/2)+1)th position towards the right and keep filling the original array starting from the 2nd position.

Below is the implementation of above idea:

C++

// C++ program to rearrange the array as per the given
// condition
 
#include <bits/stdc++.h>
using namespace std;
 
// function to rearrange the array
void rearrangeArr(int arr[], int n)
{
    // total even positions
    int evenPos = n / 2;
    // total odd positions
    int oddPos = n - evenPos;
    int tempArr[n];
   
    // copy original array in an auxiliary array
    for (int i = 0; i < n; i++)
        tempArr[i] = arr[i];
 
    // sort the auxiliary array
    sort(tempArr, tempArr + n);
    int j = oddPos - 1;
 
    // fill up odd position in original array
    for (int i = 0; i < n; i += 2)
        arr[i] = tempArr[j--];
 
    j = oddPos;
 
    // fill up even positions in original array
    for (int i = 1; i < n; i += 2)
        arr[i] = tempArr[j++];
 
    // display array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int size = sizeof(arr) / sizeof(arr[0]);
    rearrangeArr(arr, size);
    return 0;
}

C

// C program to rearrange the array as per the given
// condition
#include <stdio.h>
#include <stdlib.h>
 
// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}
 
// function to rearrange the array
void rearrangeArr(int arr[], int n)
{
    // total even positions
    int evenPos = n / 2;
    // total odd positions
    int oddPos = n - evenPos;
    int tempArr[n];
 
    // copy original array in an auxiliary array
    for (int i = 0; i < n; i++)
        tempArr[i] = arr[i];
 
    // sort the auxiliary array
    qsort(tempArr, n, sizeof(int), cmpfunc);
    int j = oddPos - 1;
 
    // fill up odd position in original array
    for (int i = 0; i < n; i += 2)
        arr[i] = tempArr[j--];
 
    j = oddPos;
 
    // fill up even positions in original array
    for (int i = 1; i < n; i += 2)
        arr[i] = tempArr[j++];
 
    // display array
    for (int i = 0; i < n; i++)
        printf("%d ", arr[i]);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int size = sizeof(arr) / sizeof(arr[0]);
    rearrangeArr(arr, size);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta

Java

// Java program to rearrange the array
// as per the given condition
import java.util.*;
import java.lang.*;
 
public class GfG{
    // function to rearrange the array
    public static void rearrangeArr(int arr[],
                                        int n)
    {
        // total even positions
        int evenPos = n / 2;
 
        // total odd positions
        int oddPos = n - evenPos;
 
        int[] tempArr = new int [n];
 
        // copy original array in an
        // auxiliary array
        for (int i = 0; i < n; i++)
            tempArr[i] = arr[i];
 
        // sort the auxiliary array
        Arrays.sort(tempArr);
 
        int j = oddPos - 1;
 
        // fill up odd position in 
        // original array
        for (int i = 0; i < n; i += 2) {
            arr[i] = tempArr[j];
            j--;
        }
 
        j = oddPos;
 
        // fill up even positions in 
        // original array
        for (int i = 1; i < n; i += 2) {
            arr[i] = tempArr[j];
            j++;
        }
 
        // display array
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
     
    // Driver function 
    public static void main(String argc[]){
        int[] arr = new int []{ 1, 2, 3, 4, 5,
                                        6, 7 };
        int size = 7;
        rearrangeArr(arr, size);
         
    }
}
 
/* This code is contributed by Sagar Shukla */

Python3

# Python3 code to rearrange the array
# as per the given condition
import array as a
import numpy as np
 
# function to rearrange the array
def rearrangeArr(arr, n):
     
    # total even positions
    evenPos = int(n / 2)
 
    # total odd positions
    oddPos = n - evenPos
 
    # initialising empty array in python
    tempArr = np.empty(n, dtype = object)
 
    # copy original array in an
    # auxiliary array
    for i in range(0, n):
         
        tempArr[i] = arr[i]
 
    # sort the auxiliary array
    tempArr.sort()
 
    j = oddPos - 1
 
    # fill up odd position in original
    # array
    for i in range(0, n, 2):
 
        arr[i] = tempArr[j]
        j = j - 1
     
    j = oddPos
 
    # fill up even positions in original
    # array
    for i in range(1, n, 2):
        arr[i] = tempArr[j]
        j = j + 1
     
    # display array
    for i in range(0, n):
        print (arr[i], end = ' ')
 
# Driver code
arr = a.array('i', [ 1, 2, 3, 4, 5, 6, 7 ])
rearrangeArr(arr, 7)
 
# This code is contributed by saloni1297

C#

// C# program to rearrange the array
// as per the given condition
using System;
 
public class GfG {
     
    // Function to rearrange the array
    public static void rearrangeArr(int []arr, int n)
    {
        // total even positions
        int evenPos = n / 2;
 
        // total odd positions
        int oddPos = n - evenPos;
 
        int[] tempArr = new int [n];
 
        // copy original array in an
        // auxiliary array
        for (int i = 0; i < n; i++)
            tempArr[i] = arr[i];
 
        // sort the auxiliary array
        Array.Sort(tempArr);
 
        int j = oddPos - 1;
 
        // Fill up odd position in 
        // original array
        for (int i = 0; i < n; i += 2) {
            arr[i] = tempArr[j];
            j--;
        }
 
        j = oddPos;
 
        // Fill up even positions in 
        // original array
        for (int i = 1; i < n; i += 2) {
            arr[i] = tempArr[j];
            j++;
        }
 
        // display array
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
     
    // Driver Code 
    public static void Main()
    {
        int[] arr = new int []{ 1, 2, 3, 4, 5, 6, 7 };
        int size = 7;
        rearrangeArr(arr, size);
    }
}
 
/* This code is contributed by vt_m */

PHP

<?php 
// PHP program to rearrange the array
// as per the given condition
 
// function to rearrange the array
function rearrangeArr(&$arr, $n)
{
    // total even positions
    $evenPos = intval($n / 2);
 
    // total odd positions
    $oddPos = $n - $evenPos;
 
    $tempArr = array_fill(0, $n, NULL);
 
    // copy original array in an
    // auxiliary array
    for ($i = 0; $i < $n; $i++)
        $tempArr[$i] = $arr[$i];
 
    // sort the auxiliary array
    sort($tempArr);
 
    $j = $oddPos - 1;
 
    // fill up odd position in 
    // original array
    for ($i = 0; $i < $n; $i += 2) 
    {
        $arr[$i] = $tempArr[$j];
        $j--;
    }
 
    $j = $oddPos;
 
    // fill up even positions in 
    // original array
    for ($i = 1; $i < $n; $i += 2)
    {
        $arr[$i] = $tempArr[$j];
        $j++;
    }
 
    // display array
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i] ." ";
}
 
// Driver code
$arr = array(1, 2, 3, 4, 5, 6, 7 );
$size = sizeof($arr);
rearrangeArr($arr, $size);
 
// This code is contributed 
// by ChitraNayal
?>

Javascript

<script>
 
 
// Javascript program to rearrange the array
// as per the given condition
 
 
// function to rearrange the array
function rearrangeArr(arr, n)
{
    // total even positions
    let evenPos = Math.floor(n / 2);
 
    // total odd positions
    let oddPos = n - evenPos;
 
    let tempArr = new Array(n);
 
    // copy original array in an
    // auxiliary array
    for (let i = 0; i < n; i++)
        tempArr[i] = arr[i];
 
    // sort the auxiliary array
    tempArr.sort();
 
    let j = oddPos - 1;
 
    // fill up odd position in original
    // array
    for (let i = 0; i < n; i += 2) {
        arr[i] = tempArr[j];
        j--;
    }
 
    j = oddPos;
 
    // fill up even positions in original
    // array
    for (let i = 1; i < n; i += 2) {
        arr[i] = tempArr[j];
        j++;
    }
 
    // display array
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Driver code
 
    let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
    let size = arr.length;
    rearrangeArr(arr, size);
    
 
//This code is contributed by Mayank Tyagi
</script>

Output

4 5 3 6 2 7 1

Time Complexity: O( n logn )
Auxiliary Space: O(n)

Another Approach-

We can traverse the array by defining two variables p and q and assign values from last.

if even index is there, we will give it max value otherwise min value.

p =0 and q= end;

p will go ahead and q will decrease.

C++

#include <bits/stdc++.h>
using namespace std;
 
int main(){
    int n,i,p,q;
    int a[]= {1, 2, 1, 4, 5, 6, 8, 8};
    n=sizeof(a)/sizeof(a[0]);
    int b[n];
    for(i=0;i<n;i++)
        b[i]=a[i];
 
    sort(b,b+n);
    p=0;q=n-1;
    for(i=n-1;i>=0;i--){
            if(i%2!=0){
            a[i]=b[q];
            q--;
            }
            else{
                a[i]=b[p];
                p++;
            }
    }
    for(i=0;i<n;i++){
        cout<<a[i]<<" ";
    }
    return 0;
}

Java

import java.util.*;
 
class GFG{
 
  public static void main(String[] args)
  {
    int n, i, j, p, q;
    int a[] = {1, 2, 1, 4, 5, 6, 8, 8};
    n = a.length;
    int []b = new int[n];
    for(i = 0; i < n; i++)
      b[i] = a[i];
 
    Arrays.sort(b);
    p = 0; q = n - 1;
    for(i = n - 1; i >= 0; i--)
    {
      if(i % 2 != 0)
      {
        a[i] = b[q];
        q--;
      }
      else{
        a[i] = b[p];
        p++;
      }
    }
    for(i = 0; i < n; i++)
    {
      System.out.print(a[i]+" ");
    }
  }
}
 
// This code is contributed by gauravrajput1 

Python3

if __name__ == '__main__':
    #n, i, j, p, q;
    a = [ 1, 2, 1, 4, 5, 6, 8, 8 ];
    n = len(a);
    b = [0]*n;
    for i in range(n):
        b[i] = a[i];
 
    b.sort();
    p = 0;
    q = n - 1;
    for i in range(n-1, -1,-1):
        if (i % 2 != 0):
            a[i] = b[q];
            q -= 1;
        else:
            a[i] = b[p];
            p += 1;
         
    for i in range(n):
        print(a[i], end=" ");
     
# This code is contributed by gauravrajput1

C#

using System;
 
 
public class GFG{
 
  public static void Main(String[] args)
  {
    int n, i, j, p, q;
    int []a = {1, 2, 1, 4, 5, 6, 8, 8};
    n = a.Length;
    int []b = new int[n];
    for(i = 0; i < n; i++)
      b[i] = a[i];
 
    Array.Sort(b);
    p = 0; q = n - 1;
    for(i = n - 1; i >= 0; i--)
    {
      if(i % 2 != 0)
      {
        a[i] = b[q];
        q--;
      }
      else{
        a[i] = b[p];
        p++;
      }
    }
    for(i = 0; i < n; i++)
    {
      Console.Write(a[i]+" ");
    }
  }
}
 
// This code is contributed by gauravrajput1

Javascript

<script>
     
        var n, i, j, p, q;
        var a = [ 1, 2, 1, 4, 5, 6, 8, 8 ];
        n = a.length;
        var b = Array(n).fill(0);
        for (i = 0; i < n; i++)
            b[i] = a[i];
 
        b.sort();
        p = 0;
        q = n - 1;
        for (i = n - 1; i >= 0; i--) {
            if (i % 2 != 0) {
                a[i] = b[q];
                q--;
            } else {
                a[i] = b[p];
                p++;
            }
        }
        for (i = 0; i < n; i++) {
            document.write(a[i] + " ");
        }
 
// This code is contributed by gauravrajput1 
</script>

Output

4 5 2 6 1 8 1 8

Time Complexity: O(n log n), The maximum time taken to sort the array.
Auxiliary Space: O(n), The extra space is required to store the copy of elements of original array.

This algorithm will take 1 for loop less than the previous one.

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